Numerical simulation of a viscous Oldroyd-B model

Transcription

1 Numerical simulation of a viscous Oldroyd-B model Bangwei She, Mária Lukáčová JGU-Mainz cooperation with Prof. M.Tabata, H. Notsu, A. Tezuka Waseda University IRTG 1529 Mathematical fluid dynamics Sino-German Symposium on advanced numerical methods for compressible fluid mechanics and related problems

2 Oldroyd-B model Re( u t +u u) = p +α u+ β We σ u = 0 (1) σ t +(u )σ u σ σ ( u)t = 1 We (I σ) u: velocity, p: pressure Re: Reynolds number σ: conformation tensor, symmetric positive definite, elastic We: Weissenberg number, relaxation time over characteristic time α: ratio of Newtonian viscocity in total viscocity. β = 1 α. B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

3 Existence results local in time existence: Guillope and Saut, 1990 Renardy 1991 Jourdain, Lelivre and Bris, 2004 Li and Zhang,2004 B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

4 Existence results local in time existence: Guillope and Saut, 1990 Renardy 1991 Jourdain, Lelivre and Bris, 2004 Li and Zhang,2004 global existence with small initial data: Guillope and Saut, 1990 Lin, Liu and Zhang, 2005 B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

5 Existence results local in time existence: Guillope and Saut, 1990 Renardy 1991 Jourdain, Lelivre and Bris, 2004 Li and Zhang,2004 global existence with small initial data: Guillope and Saut, 1990 Lin, Liu and Zhang, 2005 regularized model: regularity in 2, Constantin,2012 global existence of regularized model: Barrett, 2014 B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

6 Existence results local in time existence: Guillope and Saut, 1990 Renardy 1991 Jourdain, Lelivre and Bris, 2004 Li and Zhang,2004 global existence with small initial data: Guillope and Saut, 1990 Lin, Liu and Zhang, 2005 regularized model: regularity in 2, Constantin,2012 global existence of regularized model: Barrett, 2014 global existence and uniqueness in the discrete scheme: Lee, Xu, and Zhang, 2011 B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

7 Numerical results High Weissenberg number problem. 3.5 x Re = 1, We = B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

8 Positivity-preserving method: B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

9 Positivity-preserving method: Log-transformation: ψ = logσ Fattal and Kupfermann, 2005 B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

10 Positivity-preserving method: Log-transformation: ψ = logσ Fattal and Kupfermann, 2005 Square-root: ψ = (σ) 1 2 B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

11 Positivity-preserving method: Log-transformation: ψ = logσ Fattal and Kupfermann, 2005 Square-root: ψ = (σ) 1 2 Euler-Lagrangian method: δσ δt = σ +(u )σ u σ σ ( u)t t is discretized as σk+1 F(σ k X)F T t df dt = uf Trebotich, 2005 Lee and Xu, 2006 Lee, Xu, and Zhang, 2011 B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

12 A cartoon model: φ t +a(x) φ x b(x)φ = 1 We φ φ(x,0) = 0. a(x),b(x) > 0 play the role as u, u. Steady state φ(x) = x ) We 1 0 exp(b(x a(x ) )dx is exponential. B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

13 A cartoon model: φ t +a(x) φ x b(x)φ = 1 We φ φ(x,0) = 0. a(x),b(x) > 0 play the role as u, u. Steady state φ(x) = x ) We 1 0 exp(b(x a(x ) )dx is exponential. What would happen if we do logarithm transformation? ψ = logφ. B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

14 A cartoon model: φ t +a(x) φ x b(x)φ = 1 We φ φ(x,0) = 0. a(x),b(x) > 0 play the role as u, u. Steady state φ(x) = x ) We 1 0 exp(b(x a(x ) )dx is exponential. What would happen if we do logarithm transformation? ψ = logφ. ψ t +a(x) ψ x b(x) = 1 We B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

15 A cartoon model: φ t +a(x) φ x b(x)φ = 1 We φ φ(x,0) = 0. a(x),b(x) > 0 play the role as u, u. Steady state φ(x) = x ) We 1 0 exp(b(x a(x ) )dx is exponential. What would happen if we do logarithm transformation? ψ = logφ. ψ t +a(x) ψ x b(x) = 1 We Numerical methods based on polynomials is difficult to catch the exponential growth!! Re = 1, We = 2, t = τ y (x=0.5) B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

16 some stability technique For any matrix u and symmetric positive definite matrix σ: u = B+Ω+Nσ 1. N,Ω are anti-symmetric, B is symmetric and commutes with τ. B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

17 some stability technique For any matrix u and symmetric positive definite matrix σ: u = B+Ω+Nσ 1. N,Ω are anti-symmetric, B is symmetric and commutes with τ. Eq.(1) 3 σ t +(u )σ u σ σ ( u)t = 1 We (I σ) can be written as σ t +(u )σ = Ωσ σω+2bσ + 1 We (I σ) B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

18 some stability technique For any matrix u and symmetric positive definite matrix σ: u = B+Ω+Nσ 1. N,Ω are anti-symmetric, B is symmetric and commutes with τ. Eq.(1) 3 σ t +(u )σ u σ σ ( u)t = 1 We (I σ) can be written as σ t +(u )σ = Ωσ σω+2bσ + 1 We (I σ) Log-transformation ψ = log(σ): diag(λ i ) = R T σr,ψ = Rdiag(logλ i )R T. B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

19 some stability technique For any matrix u and symmetric positive definite matrix σ: u = B+Ω+Nσ 1. N,Ω are anti-symmetric, B is symmetric and commutes with τ. Eq.(1) 3 σ t +(u )σ u σ σ ( u)t = 1 We (I σ) can be written as σ t +(u )σ = Ωσ σω+2bσ + 1 We (I σ) Log-transformation ψ = log(σ): diag(λ i ) = R T σr,ψ = Rdiag(logλ i )R T. ψ t +u ψ = Ωψ ψω+2b+ 1 We (e ψ I) B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

20 kinetic energy We=0.5 We=1 We= time No blow up!! B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

21 kinetic energy We=0.5 We=1 We= time No blow up!! Table: L 2 error σ h σ h/2 h We = 0.5 We = 1 We = 3 1/ / / B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

22 A viscous Oldroyd-B model in log-transformation: u t +u u = p +α u+ β We eψ u = 0 ψ t +u ψ = Ωψ ψω+2b+ 1 We (e ψ I)+ε ψ (2) B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

23 A viscous Oldroyd-B model in log-transformation: u t +u u = p +α u+ β We eψ u = 0 ψ t +u ψ = Ωψ ψω+2b+ 1 We (e ψ I)+ε ψ (2) The viscous model satisfies the inequality: d dt F(u,σ)+α u 2 + β 2We 2 tr(σ +σ 1 2I) 0 (3) and the free-energy F(u, σ) decreases exponentially fast to zero in time. B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

24 A viscous Oldroyd-B model in log-transformation: u t +u u = p +α u+ β We eψ u = 0 ψ t +u ψ = Ωψ ψω+2b+ 1 We (e ψ I)+ε ψ (2) The viscous model satisfies the inequality: d dt F(u,σ)+α u 2 + β 2We 2 tr(σ +σ 1 2I) 0 (3) and the free-energy F(u, σ) decreases exponentially fast to zero in time. The free-energy F(u,σ) = Re 2 u 2 + β tr(σ logσ I) (4) 2We is found to be the entropy(hu and Lelièvre, 2006). B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

25 Inner product of the Navier-Stokes equation with the velocity: Re d u 2 +α 2 dt u 2 + β u : e ψ = 0 (5) We B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

26 Inner product of the Navier-Stokes equation with the velocity: Re d u 2 +α 2 dt u 2 + β u : e ψ = 0 (5) We For (2) 3 ψ t +u ψ = Ωψ ψω+2b+ 1 We (e ψ I)+ε ψ we contract it with e ψ : d dt ψ : eψ = d dt tr(eψ ) (Ωψ ψω) : e ψ = tr(ωψe ψ ) tr(ψωe ψ ) = 0. u : e ψ = Ω : e ψ +B : e ψ +Ne ψ : e ψ = B : e ψ (e ψ I) : e ψ = tr(i e ψ ) B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

27 Inner product of the Navier-Stokes equation with the velocity: Re d u 2 +α 2 dt u 2 + β u : e ψ = 0 (5) We For (2) 3 ψ t +u ψ = Ωψ ψω+2b+ 1 We (e ψ I)+ε ψ we contract it with e ψ : d dt ψ : eψ = d dt tr(eψ ) (Ωψ ψω) : e ψ = tr(ωψe ψ ) tr(ψωe ψ ) = 0. u : e ψ = Ω : e ψ +B : e ψ +Ne ψ : e ψ = B : e ψ (e ψ I) : e ψ = tr(i e ψ ) d tr(e ψ ) = 2 u : e ψ + 1 dt We tr(i eψ )+ε ψ : e ψ (6) B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

28 Inner product of the Navier-Stokes equation with the velocity: Re d u 2 +α 2 dt u 2 + β u : e ψ = 0 (5) We For (2) 3 ψ t +u ψ = Ωψ ψω+2b+ 1 We (e ψ I)+ε ψ we contract it with e ψ : d dt ψ : eψ = d dt tr(eψ ) (Ωψ ψω) : e ψ = tr(ωψe ψ ) tr(ψωe ψ ) = 0. u : e ψ = Ω : e ψ +B : e ψ +Ne ψ : e ψ = B : e ψ (e ψ I) : e ψ = tr(i e ψ ) d tr(e ψ ) = 2 u : e ψ + 1 dt We tr(i eψ )+ε ψ : e ψ (6) (5) + β 2We (6) B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

29 d dt + = εβ 2We ( Re 2 u 2 + β ) 2We tr(eψ I) ( α u 2 + β ) 2We 2tr(eψ I) ψ : e ψ = εβ 2We ψ : e ψ 0 (7) B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

30 d dt + = εβ 2We ( Re 2 u 2 + β ) 2We tr(eψ I) ( α u 2 + β ) 2We 2tr(eψ I) ψ : e ψ = εβ 2We e ψ x : ψ x 0. e ψ x e ψ ψ x. ψ = Rdiag(λ i )R T,σ = Rdiag(e λi )R T R = ( cosθ sinθ sinθ cosθ ψ : e ψ 0 (7) ) B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

31 d dt + = εβ 2We ( Re 2 u 2 + β ) 2We tr(eψ I) ( α u 2 + β ) 2We 2tr(eψ I) ψ : e ψ = εβ 2We e ψ x : ψ x 0. e ψ x e ψ ψ x. ψ = Rdiag(λ i )R T,σ = Rdiag(e λi )R T R = ( cosθ sinθ sinθ cosθ ψ : e ψ 0 (7) Taking the trace of (2) 3 d tr(ψ) = 1 dt We tr(e ψ I) (8) ) B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

32 d dt + = εβ 2We ( Re 2 u 2 + β ) 2We tr(eψ I) ( α u 2 + β ) 2We 2tr(eψ I) ψ : e ψ = εβ 2We e ψ x : ψ x 0. e ψ x e ψ ψ x. ψ = Rdiag(λ i )R T,σ = Rdiag(e λi )R T R = ( cosθ sinθ sinθ cosθ ψ : e ψ 0 (7) Taking the trace of (2) 3 d tr(ψ) = 1 dt We tr(e ψ I) (8) (7)- β 2We (8) B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20 )

33 d dt + 0 ( Re 2 u 2 + β ) 2We tr(eψ ψ I) ( α u 2 + β ) 2We 2tr(eψ +e ψ 2I) B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

34 d dt + 0 Poincaré inequality: ( Re 2 u 2 + β ) 2We tr(eψ ψ I) ( α u 2 + β ) 2We 2tr(eψ +e ψ 2I) u 2 C p u 2. B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

35 d dt + 0 Poincaré inequality: ( Re 2 u 2 + β ) 2We tr(eψ ψ I) ( α u 2 + β ) 2We 2tr(eψ +e ψ 2I) u 2 C p u 2. tr(e ψ +e ψ 2I) tr(e ψ ψ I) 0. B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

36 d dt + 0 Poincaré inequality: ( Re 2 u 2 + β ) 2We tr(eψ ψ I) ( α u 2 + β ) 2We 2tr(eψ +e ψ 2I) u 2 C p u 2. tr(e ψ +e ψ 2I) tr(e ψ ψ I) 0. d dt F(u,σ) cf(u,σ) 0(c = min( 2α 1, Re C p We ) > 0). Apply the Gronwall inequality, F(u,σ) F(u(t = 0),σ(t = 0))e ct (9) B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

37 characteristic method - time discretization t Let X be the position of a particle, { d dt X = u(x,t), t [tn,t n+1 ], X(t;x) = x. Material derivative: φ t = φ t +u φ is discretized as φ t φ φ(x(t t;x),t t) t t n+1 t n x=x(t,x) B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

38 characteristic method - time discretization t Let X be the position of a particle, { d dt X = u(x,t), t [tn,t n+1 ], X(t;x) = x. Material derivative: φ t = φ t +u φ is discretized as φ t φ φ(x(t t;x),t t) t t n+1 t n x=x(t,x) An implicit scheme which also satisfies the entropy inequality in the discrete level: Re uk+1 u k X(u k, t) t + p k+1 α u k+1 = β We eψk+1 u k+1 = 0 ψ k+1 ψ k X(u k, t) t = Ω k+1 ψ k+1 ψ k+1 Ω k+1 +2B k We (e ψk+1 I)+ε ψ k+1 B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

39 A characteristic finite element scheme For a given (u n h,pn h,ψn h ), find (u n+1 h,p n+1 h,ψ n+1 h ) (P1) d P1 (P1) d(d+1) 2, such that for any test function (v,q,φ) (P1) d P1 (P1) d(d+1) 2 and for all n = 0,,N T 1: Re t (uk+1 h,v)+2α((u k+1 h ),(v)) (p k+1 h, v) ( u k+1 h,q) +s h (p k+1 h,q) = Re t (uk h X,v)+ β We ( eψ 1 t (ψk+1 h,φ)+ε( ψ k+1 h, φ) = (Ω k+1 h ψ k+1 h ψ k+1 h Ω k+1 h k+1 h +( 1 t ψk h X 1(u k h, t),φ)+ 1 We (e ψk+1 h Penalty stabilization: s h (p,q) = δ κ h2 κ ( p, q) κ.,v) +2B k+1,φ) I,φ) B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

40 A characteristic finite element scheme For a given (u n h,pn h,ψn h ), find (u n+1 h,p n+1 h,ψ n+1 h ) (P1) d P1 (P1) d(d+1) 2, such that for any test function (v,q,φ) (P1) d P1 (P1) d(d+1) 2 and for all n = 0,,N T 1: Re t (uk+1 h,v)+2α((u k+1 h ),(v)) (p k+1 h, v) ( u k+1 h,q) +s h (p k+1 h,q) = Re t (uk h X,v)+ β We ( eψ 1 t (ψk+1 h,φ)+ε( ψ k+1 h, φ) = (Ω k+1 h ψ k+1 h ψ k+1 h Ω k+1 h k+1 h +( 1 t ψk h X 1(u k h, t),φ)+ 1 We (e ψk+1 h Penalty stabilization: s h (p,q) = δ κ h2 κ ( p, q) κ.,v) +2B k+1,φ) I,φ) The discrete energy inequality can be derived by taking the test function as (v,q,φ) = (u k+1 h, p k+1 h, I)) β (eψk+1 h 2 We B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

41 Numerical Test lid-driven cavity Geometry: (0,1)x(0,1) Initial values: u = (0,0), σ = I. Boundary condition for velocity: left, right and bottom, u = 0; top u = (16x 2 (1 x) 2,0). Timestep dt = hk(average size of the mesh) Re=1, We=0.5,1,3,5. α = 0.5 u B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

42 Velocity component and energy at We = 5 Velocity X Velocity Y kinetic Re=1,We=5 6 tr(σ ln(σ) I) t t B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

43 Conformation tensor and streamline at We = 5 PHI PHI PHI B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

44 8 trσ 7 6 we=0.5 we=1 we= time B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

45 u v p σ 1 σ 2 σ 3 h L 2 EOC L 2 EOC L 2 EOC L 2 EOC L 2 EOC L 2 EOC 1.27e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e Table: We=0.5 B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

46 u v p σ 1 σ 2 σ 3 h L 2 EOC L 2 EOC L 2 EOC L 2 EOC L 2 EOC L 2 EOC 1.27e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e Table: We=0.5 u v p σ 1 σ 2 σ 3 h L 2 EOC L 2 EOC L 2 EOC L 2 EOC L 2 EOC L 2 EOC 1.27e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e Table: We=5.0 Results are convergent! B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20

47 Conclusion: introduction of the model research state and challenge a viscous model based on the log-formula characteristic FEM, free energy dissipative scheme mesh convergence Thank you for your attention! B. She, M. Lukáčová (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model / 20